If $P(x)=ax^2+bx+c$ and $Q(x)=-ax^2+dx+c$, $ac$is not $0$, then prove that $P(x)\cdot Q(x)=0$ has at least 2 real roots. 
If $P(x)=ax^2+bx+c$ and $Q(x)=-ax^2+dx+c$, $ac$ is not $0$, then prove that $P(x)\cdot Q(x)=0$ has at least 2 real roots.   

$P(x)\cdot Q(x)=0$
$(ax^2+bx+c)\cdot (-ax^2+dx+c)=0$ 
Now for this to have any real roots, at least one of this equation should have real roots. This is possible when either $D_P\geq 0$ or $D_Q\geq 0$.   
Hence, to answer the question I would have to prove that it is not possible for both the inequalities $b^2-4ac\leq0$ and $d^2+4ac\leq 0$ to be true simultaneously. But how do I proceed with that?
 A: A different approach. The product polynomial has the form
$$R(x):=P(x)\cdot Q(x)=-a^2x^4+\dots+c^2.$$
Hence if $a\cdot c\not=0$ then 
$$\lim_{\pm\infty} R(x)=-\infty\quad\mbox{and}\quad R(0)=c^2>0.$$
Therefore, by continuity, there are at least two distinct zeros: one in $(-\infty, 0)$ and another in $(0,+\infty)$.
P.S. Note that the same approach works for two polynomials
$$P(x)=ax^m+\dots +c,\quad Q(x)=-ax^n+\dots +c$$
with $m,n\geq 1$ and $m+n$ an even integer.
A: Hint: if $a \ne 0$ and neither quadratic has real roots then:
$$
b^2-4ac \lt 0 \\
d^2+4ac \lt 0
$$
Adding the above gives the inequality, impossible in real numbers:
$$b^2+d^2 \lt 0$$
A: Assume $a, b, c, d$ real.
Then the discriminant of $a x^{2} + b x + c$ is $b^{2} - 4 a c$, while the discriminant of $-a x^{2} + d x + c$ is $d^{2} + 4 ac$.
Since the sum of the two discriminants is non-negative:
$$
(b^{2} - 4 a c) + (d^{2} + 4 ac) = b^{2} + d^{2} \ge 0,
$$
at least one of the two discriminants is non-negative, so at least one of the two equations has real solutions.
